Cubic Zeros Aa03E3

1. **State the problem:** Find all zeros (roots), both real and non-real, of the cubic polynomial $$f(x) = x^3 + 7x^2 + 15x + 25.$$\n\n2. **Recall the formula and approach:** To find zeros of a cubic polynomial, we can try to find at least one real root by rational root theorem or trial, then factor the polynomial and solve the quadratic factor for the remaining roots.\n\n3. **Try possible rational roots:** Possible rational roots are factors of the constant term 25 divided by factors of leading coefficient 1, i.e., $\pm1, \pm5, \pm25$.\n\n4. **Test $x = -1$:** $$f(-1) = (-1)^3 + 7(-1)^2 + 15(-1) + 25 = -1 + 7 - 15 + 25 = 16 \neq 0.$$\n\n5. **Test $x = -5$:** $$f(-5) = (-5)^3 + 7(-5)^2 + 15(-5) + 25 = -125 + 175 - 75 + 25 = 0.$$ So $x = -5$ is a root.\n\n6. **Divide polynomial by $(x + 5)$:** Use polynomial division or synthetic division:\n$$\begin{aligned} &x^3 + 7x^2 + 15x + 25 \div (x + 5) = x^2 + \cancel{2x} + \cancel{5} \text{ (with remainder 0)} \end{aligned}$$\nIntermediate step showing cancellation:\n$$\frac{x^3 + 7x^2 + 15x + 25}{\cancel{x + 5}} = x^2 + \cancel{2x} + 5$$\n\n7. **Solve quadratic $x^2 + 2x + 5 = 0$:** Use quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $a=1$, $b=2$, $c=5$.\n\nCalculate discriminant:\n$$\Delta = b^2 - 4ac = 2^2 - 4 \times 1 \times 5 = 4 - 20 = -16.$$\nSince $\Delta < 0$, roots are complex conjugates.\n\n8. **Find complex roots:**\n$$x = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i.$$\n\n9. **Final answer:** The zeros of $f(x)$ are $$\boxed{-5, -1 + 2i, -1 - 2i}.$$